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Annabel wants to make banana bread for the bake sale. She went to the market to buy bananas, and she noticed fruit flies after she returned home with her produce. The number of fruit flies grows at an exponential rate modeled by the equation f(t)=8(2)t

, where f is the number of fruit flies after t days. Define an appropriate domain for the problem if Annabel bakes the banana bread 5 days after returning from the market.(1 point)
Responses

0≤f≤5
0 less than or equal to f less than or equal to 5

0≤f≤256
0 less than or equal to f less than or equal to 256

0≤t≤256
0 less than or equal to t less than or equal to 256

0≤t≤5

11 answers

To define an appropriate domain for the problem, we need to consider the time frame in which Annabel is observing the growth of fruit flies until she bakes the banana bread.

The function \( f(t) = 8(2)^t \) represents the number of fruit flies after \( t \) days. Since Annabel bakes the banana bread 5 days after returning from the market, the value of \( t \) can range from the day she returns home (which we can consider as \( t = 0 \)) to the day she bakes the bread (which is \( t = 5 \)).

Thus, the appropriate domain for \( t \) is:

\[ 0 \leq t \leq 5 \]

This means the fruit fly population is modeled from day 0 (when the bananas were bought) to day 5 (when Annabel bakes the bread).

So, the correct response is:

0 ≤ t ≤ 5.

Callie entered an art contest in second grade and won a $1,000 scholarship. The money was invested in an account paying a 9% interest rate compounded annually. The situation can be modeled by the equation a(t)=1,000(1.09)t
, where a is the amount in the account after t years. If Callie uses the scholarship 10 years later, determine which graph accurately displays the situation.(1 point)
Responses

A decreasing curve with an arrow at the end is plotted in the first quadrant of a coordinate plane. The x-axis ranges from 0 to 12 in increments of 2 and the y-axis ranges from 0 to 1,200 in increments of 200.
Image with alt text: A decreasing curve with an arrow at the end is plotted in the first quadrant of a coordinate plane. The x-axis ranges from 0 to 12 in increments of 2 and the y-axis ranges from 0 to 1,200 in increments of 200.

An increasing curve with an arrow at the end is plotted in the first quadrant of a coordinate plane. The x-axis ranges from 0 to 10 in increments of 1 and the y-axis ranges from 0 to 2,000 in increments of 500.
Image with alt text: An increasing curve with an arrow at the end is plotted in the first quadrant of a coordinate plane. The x-axis ranges from 0 to 10 in increments of 1 and the y-axis ranges from 0 to 2,000 in increments of 500.

An increasing curve with an arrow at the end is plotted in the first quadrant of a coordinate plane. The x-axis ranges from 0 to 14 in increments of 2 and the y-axis ranges from 0 to 3,500 in increments of 500.
Image with alt text: An increasing curve with an arrow at the end is plotted in the first quadrant of a coordinate plane. The x-axis ranges from 0 to 14 in increments of 2 and the y-axis ranges from 0 to 3,500 in increments of 500.

An increasing curve with an arrow at the end is plotted in the first quadrant of a coordinate plane. The x-axis ranges from 0 to 5 in increments of 1 and the y-axis ranges from 0 to 2000 in increments of 500.
Image with alt text: An increasing curve with an arrow at the end is plotted in the first quadrant of a coordinate plane. The x-axis ranges from 0 to 5 in increments of 1 and the y-axis ranges from 0 to 2000 in increments of 500.
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To determine which graph accurately displays the situation described, we can analyze the mathematical model given:

The amount in Callie's account after \( t \) years is modeled by the equation: \[ a(t) = 1,000(1.09)^t \]

This equation represents an exponential growth function because the amount in the account increases due to the interest being compounded annually at a rate of 9%.

  1. Increasing Function: Since the interest is compounded, the function \( a(t) \) will always be increasing over time.
  2. Value at \( t = 10 \): We can calculate \( a(10) \) to get an idea of expected values at that point: \[ a(10) = 1,000(1.09)^{10} \approx 1,000 \times 2.36736 \approx 2367.36 \] Hence, after 10 years, Callie can expect to have about $2,367.36 in the account.

Given this information, we can evaluate the graph options:

  • The x-axis should cover at least from 0 to 10 years (to show up to when she uses the funds).
  • The y-axis should extend beyond $2,367.36 (to account for the amount in the account).

Now, analyzing the options:

  1. First option: Decreasing curve (incorrect, should be increasing).
  2. Second option: Increasing curve, x-axis up to 10 and y-axis up to 2,000. (Does not quite cover the estimated balance after 10 years.)
  3. Third option: Increasing curve, x-axis goes to 14 and y-axis up to 3,500. (This could be a good candidate as it reaches past the amount after 10 years.)
  4. Fourth option: Increasing curve, x-axis goes only to 5 (too short to see 10 years).

Among these, the third option has an increasing curve, includes values beyond what would be expected at \( t = 10 \), and is within the reasonable range of the investment growth.

Thus, the correct graph is:

An increasing curve with an arrow at the end is plotted in the first quadrant of a coordinate plane. The x-axis ranges from 0 to 14 in increments of 2 and the y-axis ranges from 0 to 3,500 in increments of 500.

A,B,C or D

Based on the analysis of the problem:

The correct choice is:

C: An increasing curve with an arrow at the end is plotted in the first quadrant of a coordinate plane. The x-axis ranges from 0 to 14 in increments of 2 and the y-axis ranges from 0 to 3,500 in increments of 500.

This option accurately represents the exponential growth of Callie's scholarship investment over time, allowing for values past what she would have after 10 years.

Use the table to answer the question.

0 350
1
2
3
4
5
A certain population of beetles is increasing at a rate of 22% per month. If there are currently 350 beetles, the equation f(x)=350(1.22)x
can be used to model the situation. Finish the chart and plot the points to determine which graph correctly displays the situation.(1 point)
Responses

A decreasing curve with an arrow at the end is plotted on a coordinate plane. The x-axis for months ranges from 0 to 10 in increments of 1. The y-axis for Beetles ranges from 0 to 350 in increments of 50.
Image with alt text: A decreasing curve with an arrow at the end is plotted on a coordinate plane. The x-axis for months ranges from 0 to 10 in increments of 1. The y-axis for Beetles ranges from 0 to 350 in increments of 50.

A decreasing curve with an arrow at the end is plotted on a coordinate plane. The x-axis for months ranges from 0 to 10 in increments of 1. The y-axis for Beetles ranges from 0 to 350 in increments of 50.
Image with alt text: A decreasing curve with an arrow at the end is plotted on a coordinate plane. The x-axis for months ranges from 0 to 10 in increments of 1. The y-axis for Beetles ranges from 0 to 350 in increments of 50.

An increasing curve with an arrow on the top is plotted on a coordinate plane. The x-axis for months ranges from 0 to 2,000 in increments of 200. The y-axis for Beetles ranges from 0 to 10 in increments of 1.
Image with alt text: An increasing curve with an arrow on the top is plotted on a coordinate plane. The x-axis for months ranges from 0 to 2,000 in increments of 200. The y-axis for Beetles ranges from 0 to 10 in increments of 1.

An increasing curve with an arrow on the top is plotted on a coordinate plane. The x-axis for months ranges from 0 to 10 in increments of 1. The y-axis for Beetles ranges from 0 to 2000 in increments of 250.
Image with alt text: An increasing curve with an arrow on the top is plotted on a coordinate plane. The x-axis for months ranges from 0 to 10 in increments of 1. The y-axis for Beetles ranges from 0 to 2000 in increments of 250.
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To complete the table and identify which graph correctly displays the population of beetles, we will use the provided equation:

\[ f(x) = 350(1.22)^x \]

Here, \( x \) represents the number of months, and \( f(x) \) represents the population of beetles at that time.

Let's calculate the values for \( f(x) \) for \( x = 0, 1, 2, 3, 4, \) and \( 5 \):

  1. For \( x = 0 \): \[ f(0) = 350(1.22)^0 = 350(1) = 350 \]

  2. For \( x = 1 \): \[ f(1) = 350(1.22)^1 = 350(1.22) \approx 427 \]

  3. For \( x = 2 \): \[ f(2) = 350(1.22)^2 = 350(1.4884) \approx 520.94 \approx 521 \]

  4. For \( x = 3 \): \[ f(3) = 350(1.22)^3 = 350(1.804928) \approx 631.73 \approx 632 \]

  5. For \( x = 4 \): \[ f(4) = 350(1.22)^4 = 350(2.20968864) \approx 772.4 \approx 772 \]

  6. For \( x = 5 \): \[ f(5) = 350(1.22)^5 = 350(2.67482) \approx 935.19 \approx 935 \]

Now we can fill out the completed table based on these calculations:

| \( x \) (Months) | \( f(x) \) (Beetles) | |------------------|----------------------| | 0 | 350 | | 1 | 427 | | 2 | 521 | | 3 | 632 | | 4 | 772 | | 5 | 935 |

Analyzing the Graph Options:

Based on the model we have determined that the beetle population is increasing over time. Therefore, we can rule out any graphs that suggest a decreasing trend.

  1. First and Second options: Decreasing curves (incorrect, as population is increasing).
  2. Third option: An increasing curve but with the x-axis ranging from 0 to 2000 (not suitable for the problem's time frame).
  3. Fourth option: An increasing curve with an appropriate x-axis from 0 to 10 and a y-axis increasing up to 2000. This option correctly represents an increasing population of beetles over the relevant time frame.

Conclusion:

The correct graph that displays the situation is:

An increasing curve with an arrow on the top is plotted on a coordinate plane. The x-axis for months ranges from 0 to 10 in increments of 1. The y-axis for Beetles ranges from 0 to 2000 in increments of 250.

A,B,C or D.

Based on the calculations and the characteristics of the population growth we derived:

The correct choice is:

D: An increasing curve with an arrow on the top is plotted on a coordinate plane. The x-axis for months ranges from 0 to 10 in increments of 1. The y-axis for Beetles ranges from 0 to 2000 in increments of 250.

This option accurately represents the increasing beetle population based on the given rate.

Use the image to answer the question.

A decreasing curve with an arrow on the end is plotted on a coordinate plane. The curve begins at the point left parenthesis 0 comma 700 right parenthesis.

The value of a cell phone decreases exponentially at a rate of 24% each year after its release date. If the initial value of the phone is $700, the situation can be modeled by the equation v(t)=700(0.76)t
, where v is the value of the cell phone t years since its release date. Interpret the graph of the equation and identify the true statement.

(1 point)
Responses

As t increases, v decreases slowly at first and then quickly.
As t increases, v decreases slowly at first and then quickly.

As t increases, v increases quickly at first and then slowly.
As t increases, v increases quickly at first and then slowly.

As t increases, v decreases quickly at first and then slowly.
As t increases, v decreases quickly at first and then slowly.

As t increases, v increases slowly at first and then quickly.
As t increases, v increases slowly at first and then quickly.
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To interpret the graph represented by the equation \( v(t) = 700(0.76)^t \), we need to understand how an exponential decay function behaves.

  1. Exponential Decay: The function models the value of a cell phone that decreases in value at a rate of 24% per year. This means that each year, the phone retains only 76% of its value from the previous year (because \( 1 - 0.24 = 0.76 \)).

  2. Graph Behavior:

    • Initially, right after the phone is released at \( t = 0 \), its value is at its maximum ($700).
    • As \( t \) increases, the value \( v(t) \) decreases each year. However, the rate of decrease is not constant.
    • In the early years, the absolute decrease in value is larger because the percentage is applied to the initial higher value. As time goes on and the value gets smaller, the actual dollar amount lost each year becomes less and less, causing the curve to flatten out.

  3. Interpretation of the Statements:

    • "As t increases, v decreases slowly at first and then quickly." – Incorrect (The opposite is true: it decreases quickly at first, then slowly.)
    • "As t increases, v increases quickly at first and then slowly." – Incorrect (The value does not increase; it decreases.)
    • "As t increases, v decreases quickly at first and then slowly." – Correct (This accurately describes the behavior of an exponential decay.)
    • "As t increases, v increases slowly at first and then quickly." – Incorrect (Again, the value does not increase; it decreases.)

Conclusion:

The true statement is:

As t increases, v decreases quickly at first and then slowly.